I apologize, this is my third correction to my answer. This question is very subtle

indeed. I hope this answer is the ultimate one!

First of all, if you want to take advantage of Lie's theorem

you mention (some time called third Lie theorem), the Lie algebra has to

be real, as it must be the Lie algebra of a real Lie group. Then, if you

are interested in quantum mechanics applications, I mean if you wish

that the given generators are also generators of a unitary

representation of a Lie group, the generators must be Hermitian at least

and a, a† are not.

So you first have to pass to anti self-adjoint generators (*), for

instance, introducing two constants $\omega, m >0$:

$$-iI,-iH,-iP, -iK:= -imX\qquad (1)$$

where, up to real factors (so without changing the real Lie algebra) X

and P are given by $a+a^\dagger$ and $i(a−a^\dagger)$ as is well known.

$m$ has the physical meaning of mass of the particle and $H = \hbar

\omega(a^\dagger a + \frac{1}{2}I)$ can be re-arranged to:

$$H = \frac{1}{2m}P^2 + \frac{m\omega^2}{2}X^2$$

Operators (1) are, in fact essentially self-adjoint on the dense set

made of finite linear combinations of vectors $|n\rangle$, eigenstates

of $H$.

Classically, the Galileo group in one dimension includes time

translations, space translations along x, Galileian boosts along x.

If we think of the point $(x,p)$ in the space of phases as the

vector

$(1,1,x,p) \in \mathbb R^4$, and the generic element of $G$ is denoted

by a triple $(\tau,a,v)$ (time translation + space translation + boost)

$G$ acts on the system as

$$(1,1, x, v)^t \mapsto A(\tau,a,v) (1,1, x, p)^t$$

where, for the harmonic oscillator system $A(\tau, a,v)$ is (barring

errors in computations) the $4\times 4$ real matrix

$$A(t,a,v) = \begin{bmatrix}

I & 0 \\

R_{t}T_{a,v}& R_t

\end{bmatrix} $$

where $R_t$ and $T_{a,v}$ are $2\times 2$ respectively matrices defined as:

$$R_t = \begin{bmatrix}

\cos \omega t & -\frac{\sin \omega t}{m\omega} \\

m\omega \sin \omega t & \cos \omega t

\end{bmatrix} $$

and

$$T_{a,v} = \begin{bmatrix}

a & 0 \\

0 & mv

\end{bmatrix} $$

In this way, we have $3$ generators $h, \pi, k$ obtaining by taking the

derivative of $A(t,a,v)$ respectively in $t$, $a$ and $v$ at (0,0,0).

The commutation relations of these generators are the same as for

$$H,P,K$$

with the following exception:

$$[\pi, k]=0\quad \mbox{instead of}\quad [\pi, k] = m$$

to be compared with:

$$[-iP,-iK]= - m(-iI)$$

Notice that this commutator is just a number, so that, when you

exponentiate the generators it gives rise to a phase which commutes with

all operators.

In other words, if you wish to construct an unitary representation of

$G$ acting in the Hilbert space of the harmonic oscillator, you face a

problem with the composition rule, as you find a so-called

unitary-projective representation:

$$U(g)U(g')= e^{i\alpha(g,g')}U_{gg'}\qquad (2)$$

The phase $e^{i\alpha(g,g')}$ arises when $g$ and $g'$ includes

transformations generated by the momentum $P$ and the boost $K$.

It is possible to compute $\alpha(g,g')$ using several procedures, e.g.

Hausdorff-Campbell-ecc... identity.

Notice that the mass $m$ explicitly shows up in $\alpha$ (which has just the form $\alpha(g,g')= m f(g,g')$) and this is related to

Bargmann's superselection rule.

To obtain a true unitary representation of some Lie group one can deals

with as follows. Start from the group $U(1) \times G$ (a so called

central extension of $G$) with the composition rule:

$$(e^{ia}, g) \circ (e^{ia'}, g') = (e^{i(a+a'+ \alpha(g,g'))}, gg') $$

and define the map:

$$U(1)\times G \ni (e^{ia}, g) \mapsto V_{(e^{ia}, g)} := e^{ia}U_g\:.$$

Just in view of (2), this is a proper unitary representation of

$U(1)\times G$.

Notice that $U(1)\times G$ has now a further generator commuting with

all the other generators in view of the fact that we have ``added''

$U(1)$ to the initial group $G$. This generator, in the Hilbert space,

is proportional to $-iI$. The anti-self-adjoint generators are just:

$$-iI,-iH,-iP, -iK\:.$$

So, we can conclude that the considered generators are a representation

of the Lie algebra of a central extension of a group $G$ representing the action of Galileo group along the *x* axis on the harmonic oscillator.

There are some open issues.

(1) $U(1) \times G$ is a Lie group. What is the differential structure

but also the topology on it? This is a delicate problem solved by Wigner.

(2) In view of the commutation relations of $H$ and $P$, the latter is not a conserved

quantity along time evolution. This is a consequence of the fact that the system, obviously, is not invariant under

space translations (the location of the minimum value of the harmonic

potential fixes a natural origin). Nevertheless the system admits a

conserved quantity associated with the generator $P$.

Since $-iP$ belongs the the Lie algebra of the representation,

$$e^{-itH} (-iP) e^{itH}$$

still belongs to that Lie algebra in view of the fact that $ e^{-itH}$

is a one-parameter subgroup of the representation. As a matter of fact

(barring trivial errors in computations)

$$e^{-itH} P e^{itH} = -\frac{\omega\sin (\omega t)}{m} K + \cos(\omega

t) P\:.$$

Therefore the explicitly depending on time observable in Schroedinger

picture:

$$P(t) := -\frac{\omega\sin (\omega t)}{m} K + \cos(\omega t) P$$

turns out to be a constant in Heisenberg picture:

$$P(t)_H = e^{itH} P(t) e^{-itH} = P\:.$$

This is exactly the procedure exploited to associate a constant quantity

(always in Heisenberg picture) to the boost generator, even in

relativistic theories.

-------------------------

(*) When one unitarily represents Lie groups, the Lie algebra of the

group is isomorphic to the corresponding Lie algebra of anti self-adjoint generators of the unitary representation. It is true when identifying the Lie algebra commutator with the operator commutator.